Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
double dp[2][3001]={},ar[3001],to=0;
int main()
{
	int n;cin>>n;
	for (int i=0;i<n;i++) {cin>>ar[i];}
	dp[0][0]=1;
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<=n;j++)
		{
			dp[i%2][j] = (dp[(i-1)%2][j-1] * ar[i-1]) + (dp[(i-1)%2][j] * (1-ar[i-1])); 
		}
	}
	for (int j=n/2+1;j<=n;j++)
	{
		to+=dp[n%2][j];
	}
	cout<<setprecision(9)<<to<<endl;
	//printf("%.9lf\n",to);
	return 0;
}